Chlorine-34 has a half-life of only 1.53 sec. How long does it take for 99.99% of a $^{34} C l$ sample to decay?

t = 20.34 s e c

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t = 30.34 s e c

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t = 40.34 s e c

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None of these

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Solution

The correct option is A $t = 20.34 s e c$
Let initial amount $= 100 N_{0}$

Amount decomposed = 99.99%

N = amount left $= 100 - 99.99$

$= 0.01 N_{0}$

$∴ λ = \frac{2.303}{t} l o g \frac{N_{0}}{N}$
$\frac{0.693}{1.53} = \frac{2.303}{t} l o g \frac{100}{0.01}$
$t = 20.34 s e c$

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Chlorine-34 has a half-life of only 1.53 sec. How long does it take for 99.99% of a 34Cl sample to decay?

The correct option is A t=20.34 secLet initial amount =100N0Amount decomposed = 99.99%N = amount left =100−99.99 =0.01N0 ∴λ=2.303tlogN0N 0.6931.53=2.303tlog1000.01 t=20.34 sec

Chlorine-34 has a half-life of only 1.53 sec. How long does it take for 99.99% of a $^{34} C l$ sample to decay?

The correct option is A $t = 20.34 s e c$
Let initial amount $= 100 N_{0}$

Amount decomposed = 99.99%

N = amount left $= 100 - 99.99$

$= 0.01 N_{0}$

$∴ λ = \frac{2.303}{t} l o g \frac{N_{0}}{N}$
$\frac{0.693}{1.53} = \frac{2.303}{t} l o g \frac{100}{0.01}$
$t = 20.34 s e c$